gaokao

2019 national-I-science_gkztc

18 maths questions

Q3 Inequalities Set Operations View
3. After the examination ends, submit both this test paper and the answer sheet. Section I: Multiple Choice Questions: This section has 12 questions, each worth 5 points, for a total of 60 points. For each question, only one of the four options is correct.
1. Given sets $M = \{ x \mid - 4 < x < 2 \} , N = \left\{ x \mid x ^ { 2 } - x - 6 < 0 \right\}$ , then $M \cap N =$
A. $\{ x \mid - 4 < x < 3 \}$
B. $\{ x \mid - 4 < x < - 2 \}$
C. $\{ x \mid - 2 < x < 2 \}$
D. $\{ x \mid 2 < x < 3 \}$
2. Let complex number $z$ satisfy $| z - \mathrm { i } | = 1$ , and the point corresponding to $z$ in the complex plane is $( x , y )$ , then
A. $( x + 1 ) ^ { 2 } + y ^ { 2 } = 1$
B. $( x - 1 ) ^ { 2 } + y ^ { 2 } = 1$
C. $x ^ { 2 } + ( y - 1 ) ^ { 2 } = 1$
D. $x ^ { 2 } + ( y + 1 ) ^ { 2 } = 1$
3. Given $a = \log _ { 2 } 0.2 , b = 2 ^ { 0.2 } , c = 0.2 ^ { 0.3 }$ , then
A. $a < b < c$
B. $a < c < b$
C. $c < a < b$
D. $b < c < a$
5. The graph of function $f ( x ) = \frac { \sin x + x } { \cos x + x ^ { 2 } }$ on $[ - \pi , \pi ]$ is approximately
A. [Figure]
B. [Figure]
C. [Figure]
D. [Figure]
Q6 Combinations & Selection Combinatorial Probability View
6. In ancient Chinese classics, the ``Book of Changes'' uses ``hexagrams'' to describe the changes of all things. Each ``hexagram'' consists of 6 lines arranged from bottom to top, with lines divided into yang lines ``——'' and yin lines ``——'', as shown in the figure. If a hexagram is randomly selected from all hexagrams, the probability that it has exactly 3 yang lines is
A. $\frac { 5 } { 16 }$
B. $\frac { 11 } { 32 }$
C. $\frac { 21 } { 32 }$
D. $\frac { 11 } { 16 }$
Q7 Vectors Introduction & 2D Angle or Cosine Between Vectors View
7. Given non-zero vectors $\boldsymbol { a } , \boldsymbol { b }$ satisfying $| \boldsymbol { a } | = 2 | \boldsymbol { b } |$ and $( \boldsymbol { a } - \boldsymbol { b } ) \perp \boldsymbol { b }$ , the angle between $\boldsymbol { a }$ and $\boldsymbol { b }$ is
A. $\frac { \pi } { 6 }$
B. $\frac { \pi } { 3 }$
C. $\frac { 2 \pi } { 3 }$
D. $\frac { 5 \pi } { 6 }$
Q8 Fixed Point Iteration View
8. The figure shows a flowchart for computing $\frac { 1 } { 2 + \frac { 1 } { 2 + \frac { 1 } { 2 } } }$ . The blank box should be filled with [Figure]
A. $A = \frac { 1 } { 2 + A }$
B. $A = 2 + \frac { 1 } { A }$
C. $A = \frac { 1 } { 1 + 2 A }$
D. $A = 1 + \frac { 1 } { 2 A }$
Q9 Arithmetic Sequences and Series Find General Term Formula View
9. Let $S _ { n }$ denote the sum of the first $n$ terms of an arithmetic sequence $\left\{ a _ { n } \right\}$ . Given $S _ { 4 } = 0 , a _ { 5 } = 5$ , then
A. $a _ { n } = 2 n - 5$
B. $a _ { n } = 3 n - 10$
C. $S _ { n } = 2 n ^ { 2 } - 8 n$
D. $S _ { n } = \frac { 1 } { 2 } n ^ { 2 } - 2 n$
10. Given that the foci of ellipse $C$ are $F _ { 1 } ( - 1,0 ) , F _ { 2 } ( 1,0 )$ , and a line through $F _ { 2 }$ intersects $C$ at points $A , B$ . If $\left| A F _ { 2 } \right| = 2 \left| F _ { 2 } B \right|$ and $| A B | = \left| B F _ { 1 } \right|$ , then the equation of $C$ is
A. $\frac { x ^ { 2 } } { 2 } + y ^ { 2 } = 1$
B. $\frac { x ^ { 2 } } { 3 } + \frac { y ^ { 2 } } { 2 } = 1$
C. $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 3 } = 1$
D. $\frac { x ^ { 2 } } { 5 } + \frac { y ^ { 2 } } { 4 } = 1$
11. Regarding the function $f ( x ) = \sin | x | + | \sin x |$ , there are four conclusions:
(1) $f ( x )$ is an even function
(2) $f ( x )$ is monotonically increasing on the interval $\left( \frac { \pi } { 2 } , \pi \right)$
(3) $f ( x )$ has 4 zeros on $[ - \pi , \pi ]$
(4) The maximum value of $f ( x )$ is 2
The numbers of all correct conclusions are
A. (1)(2)(4)
B. (2)(4)
C. (1)(4)
D. (1)(3)
12. Given that the four vertices of tetrahedron $P - A B C$ lie on the surface of sphere $O$ , with $P A = P B = P C$ , $\triangle A B C$ is an equilateral triangle with side length 2, $E , F$ are the midpoints of $P A , A B$ respectively, and $\angle C E F = 90 ^ { \circ }$ , then the volume of sphere $O$ is
A. $8 \sqrt { 6 } \pi$
B. $4 \sqrt { 6 } \pi$
C. $2 \sqrt { 6 } \pi$
D. $\sqrt { 6 } \pi$
Section II: Fill-in-the-Blank Questions: This section has 4 questions, each worth 5 points, for a total of 20 points.
Q13 Tangents, normals and gradients Chain Rule with Composition of Explicit Functions View
13. The equation of the tangent line to the curve $y = 3 \left( x ^ { 2 } + x \right) \mathrm { e } ^ { x }$ at the point $( 0,0 )$ is $\_\_\_\_$ .
Q14 Geometric Sequences and Series Finite Geometric Sum and Term Relationships View
14. Let $S _ { n }$ denote the sum of the first $n$ terms of a geometric sequence $\left\{ a _ { n } \right\}$ . If $a _ { 1 } = \frac { 1 } { 3 } , a _ { 4 } ^ { 2 } = a _ { 6 }$ , then $S _ { 5 } =$ $\_\_\_\_$ .
15. Teams A and B are in a basketball championship series using a best-of-seven format (the series ends when one team wins four games). Based on previous results, Team A's home/away schedule is ``home, home, away, away, home, away, home'' in order. Team A's probability of winning at home is 0.6, and away is 0.5. Each game is independent. The probability that Team A wins 4-1 is $\_\_\_\_$ .
16. Given hyperbola $C : \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > 0 , b > 0 )$ with left and right foci $F _ { 1 } , F _ { 2 }$ respectively. A line through $F _ { 1 }$ intersects the two asymptotes of $C$ at points $A , B$ respectively. If $\overrightarrow { F _ { 1 } A } = \overrightarrow { A B }$ and $\overrightarrow { F _ { 1 } B } \cdot \overrightarrow { F _ { 2 } B } = 0$ , then the eccentricity of $C$ is $\_\_\_\_$ .
Section III: Solution Questions: Total 70 points. Show all work, proofs, and calculations. Questions 17-21 are required for all students. Questions 22 and 23 are optional; choose one to answer.
(I) Required Questions: Total 60 points
Q17 12 marks Trig Proofs Triangle Trigonometric Relation View
17. (12 points) In $\triangle A B C$ , let the sides opposite to angles $A , B , C$ be $a , b , c$ respectively. Given $( \sin B - \sin C ) ^ { 2 } = \sin ^ { 2 } A - \sin B \sin C$ .
(1) Find $A$ ;
(2) If $\sqrt { 2 } a + b = 2 c$ , find $\sin C$ .
Q18 12 marks Vectors 3D & Lines Multi-Part 3D Geometry Problem View
18. (12 points) As shown in the figure, the right prism $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$ has a rhombus base, with $A A _ { 1 } = 4 , A B = 2 , \angle B A D = 60 ^ { \circ }$ . Let $E , M , N$ be the midpoints of $B C$ , $B B _ { 1 Therefore $f(x)$ has a unique zero point on $\left[\frac{\pi}{2}, \pi\right]$.
(iv) When $x \in (\pi, +\infty)$, $\ln(x+1) > 1$, so $f(x) < 0$, thus $f(x)$ has no zero points on $(\pi, +\infty)$.
In conclusion, $f(x)$ has exactly 2 zero points.
Q21 Discrete Probability Distributions Recurrence Relations and Sequences Involving Probabilities View
21. Solution: The possible values of $X$ are $-1, 0, 1$.
$$\begin{aligned} & P(X = -1) = (1-\alpha)\beta, \\ & P(X = 0) = \alpha\beta + (1-\alpha)(1-\beta), \\ & P(X = 1) = \alpha(1-\beta), \end{aligned}$$
Therefore the distribution table of $X$ is
$X$$-1$$0$$1$
$P$$(1-\alpha)\beta$$\alpha\beta + (1-\alpha)(1-\beta)$$\alpha(1-\beta)$

(2) (i) From (1) we have $a = 0.4$, $b = 0.5$, $c = 0.1$. Therefore $p_i = 0.4p_{i-1} + 0.5p_i + 0.1p_{i+1}$, so $0.1(p_{i+1} - p_i) = 0.4(p_i - p_{i-1})$, that is $p_{i+1} - p_i = 4(p_i - p_{i-1})$. Since $p_1 - p_0 = p_1 \neq 0$, the sequence $\{p_{i+1} - p_i\}$ $(i = 0, 1, 2, \cdots, 7)$ is a geometric sequence with common ratio 4 and first term $p_1$.
(ii) From (i) we obtain $p_8 = p_8 - p_7 + p_7 - p_6 + \cdots + p_1 - p_0 + p_0 = (p_8 - p_7) + (p_7 - p_6) + \cdots + (p_1 - p_0) = \frac{4^8 - 1}{3}p_1$. Since $p_8 = 1$, we have $p_1 = \frac{3}{4^8 - 1}$, therefore $p_4 = (p_4 - p_3) + (p_3 - p_2) + (p_2 - p_1) + (p_1 - p_0) = \frac{4^4 - 1}{3}p_1 = \frac{1}{257}$. $p_4$ represents the probability of ultimately concluding that drug A is more effective. From the calculation result, we can see that when drug A has a cure rate of 0.5 and drug B has a cure rate of 0.8, the probability of concluding that drug A is more effective is $p_4 = \frac{1}{257} \approx 0.0039$. The probability of reaching an incorrect conclusion is very small, indicating that this experimental scheme is reasonable.
Q22 Parametric curves and Cartesian conversion View
22. Solution: (1) Since $-1 < \frac{1-t^2}{1+t^2} \leq 1$ and $x^2 + \left(\frac{y}{2}\right)^2 = \left(\frac{1-t^2}{1+t^2}\right)^2 + \frac{4t^2}{(1+t^2)^2} = 1$, the rectangular coordinate equation of $C$ is $x^2 + \frac{y^2}{4} = 1$ $(x \neq -1)$. The rectangular coordinate equation of $l$ is $2x + \sqrt{3}y + 11 = 0$.
(2) From (1) we can set the parametric equation of $C$ as $\left\{\begin{array}{l} x = \cos\alpha, \\ y = 2\sin\alpha \end{array}\right.$ ($\alpha$ is the parameter, $-\pi < \alpha < \pi$). The distance from a point on $C$ to $l$ is $\frac{|2\cos\alpha + 2\sqrt{3}\sin\alpha + 11|}{\sqrt{7}} = \frac{4\cos\left(\alpha - \frac{\pi}{3}\right) + 11}{\sqrt{7}}$.
When $\alpha = -\frac{2\pi}{3}$, $4\cos\left(\alpha - \frac{\pi}{3}\right) + 11$ attains its minimum value of 7, therefore the minimum distance from a point on $C$ to $l$ is $\sqrt{7}$.
Q23 3 marks Proof Direct Proof of an Inequality View
23. Solution: (1) Since $a^2 + b^2 \geq 2ab$, $b^2 + c^2 \geq 2bc$, $c^2 + a^2 \geq 2ac$, and $abc = 1$, we have $a^2 + b^2 + c^2 \geq ab + bc + ca = \frac{ab + bc + ca}{abc} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$. Therefore $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq a^2 + b^2 + c^2$.
(2) Since $a, b, c$ are positive numbers and $abc = 1$, we have $(a+b)^3 + (b+c)^3 + (c+a)^3 \geq 3\sqrt[3]{(a+b)^3(b+c)^3(a+c)^3}$ $= 3(a+b)(b+c)(a+c)$ $\geq 3 \times (2\sqrt{ab}) \times (2\sqrt{bc}) \times (2\sqrt{ac})$ $= 24$. Therefore $(a+b)^3 + (b+c)^3 + (c+a)^3 \geq 24$.